Answer
The maximum value of $f\left( {x,y,z} \right) = xyz$ subject to the constraint $g\left( {x,y,z} \right) = 2x + y + 4z = 1$ is $\frac{1}{{216}}$.
Work Step by Step
We have $f\left( {x,y,z} \right) = xyz$ and the constraint $g\left( {x,y,z} \right) = 2x + y + 4z = 1$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {yz,xz,xy} \right) = \lambda \left( {2,1,4} \right)$
(1) ${\ \ \ \ }$ $yz = 2\lambda $, ${\ \ \ }$ $xz = \lambda $, ${\ \ \ }$ $xy = 4\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we have the following cases:
Case 1. $x=0$, $y=0$, $z \ne 0$. In this case $\lambda = 0$.
Substituting $x=0$, $y=0$ in the constraint $g\left( {x,y,z} \right) = 2x + y + 4z - 1 = 0$ gives $z = \frac{1}{4}$.
So, the critical point is $\left( {0,0,\frac{1}{4}} \right)$.
Case 2. $x=0$, $y \ne 0$, $z=0$. In this case $\lambda = 0$.
Substituting $x=0$, $z=0$ in the constraint $g\left( {x,y,z} \right) = 2x + y + 4z - 1 = 0$ gives $y=1$.
So, the critical point is $\left( {0,1,0} \right)$.
Case 3. $x \ne 0$, $y=0$, $z=0$. In this case $\lambda = 0$.
Substituting $y=0$, $z=0$ in the constraint $g\left( {x,y,z} \right) = 2x + y + 4z - 1 = 0$ gives $x = \frac{1}{2}$.
So, the critical point is $\left( {\frac{1}{2},0,0} \right)$.
Case 4. $x \ne 0$, $y \ne 0$, $z \ne 0$. In this case $\lambda \ne 0$.
From equation (1), we obtain $\lambda = \frac{{yz}}{2} = xz = \frac{{xy}}{4}$.
So, we have
$\frac{{yz}}{2} = xz$, ${\ \ \ \ }$ $xz = \frac{{xy}}{4}$
The first and the second equation give $y=2x$ and $z = \frac{1}{4}y$.
Substituting $y=2x$ in $z = \frac{1}{4}y$ gives $z = \frac{1}{2}x$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y=2x$ and $z = \frac{1}{2}x$. Substituting these in the constraint $g\left( {x,y,z} \right)$ gives
$2x + y + 4z - 1 = 0$
$2x + 2x + 2x = 1$
$x = \frac{1}{6}$
Using $y=2x$ and $z = \frac{1}{2}x$, we obtain the critical point: $\left( {\frac{1}{6},\frac{1}{3},\frac{1}{{12}}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Criticalpoints}}}&{f\left( {x,y,z} \right)}\\
{\left( {0,0,\frac{1}{4}} \right)}&0\\
{\left( {0,1,0} \right)}&0\\
{\left( {\frac{1}{2},0,0} \right)}&0\\
{\left( {\frac{1}{6},\frac{1}{3},\frac{1}{{12}}} \right)}&{\frac{1}{{216}}}
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y,z} \right)$ on the constraint is $\frac{1}{{216}}$ and the minimum value is $0$.
